STEM Problem Detail

Problem Statement

Prompt

Provide stepwise solution for the given Exercise 1.5#10

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**Problem 1.5#10:** Consider $$x' = \dfrac{x^2}{t^2 + 1} - 1$$.

### (a) Isoclines analysis

**Step 1: Finding isoclines**
Isoclines are curves where $$x' = m$$ (constant):
\[
\frac{x^2}{t^2 + 1} - 1 = m \quad \Rightarrow \quad x^2 = (m + 1)(t^2 + 1)
\]

**Step 2: Nature of isoclines**
For fixed $$m$$, $$x^2 = (m+1)(t^2 + 1)$$ represents:

- If $$m+1 > 0$$: Two branches $$x = \pm \sqrt{m+1} \sqrt{t^2 + 1}$$ (hyperbola-like)
- If $$m+1 = 0$$: $$x = 0$$ (line)
- If $$m+1 < 0$$: No real isocline (since $$x^2 \ge 0$$ but RHS $$< 0$$)

**Step 3: Isocline for slope $$m=0$$**
\[
x^2 = (0+1)(t^2 + 1) = t^2 + 1 \quad \Rightarrow \quad x = \pm \sqrt{t^2 + 1}
\]
These are two symmetric curves.

**Step 4: Isocline for slope $$m=-1$$**
\[
x^2 = (-1+1)(t^2 + 1) = 0 \quad \Rightarrow \quad x = 0
\]
This is the $$t$$-axis.

**Step 5: Regions of positive/negative slope**
Slope $$x' > 0$$ when:
\[
\frac{x^2}{t^2 + 1} - 1 > 0 \quad \Rightarrow \quad x^2 > t^2 + 1
\]
This occurs outside the $$m=0$$ isoclines: $$|x| > \sqrt{t^2 + 1}$$.

Slope $$x' < 0$$ when:
\[
\frac{x^2}{t^2 + 1} - 1 < 0 \quad \Rightarrow \quad x^2 < t^2 + 1
\]
This occurs inside the $$m=0$$ isoclines: $$|x| < \sqrt{t^2 + 1}$$.

### (b) Global existence for $$|u(0)| < 1$$

**Step 1: Rewrite equation**
$$x' = \dfrac{x^2}{t^2+1} - 1 = \dfrac{x^2 - (t^2+1)}{t^2+1}$$

**Step 2: Show solutions stay bounded**
Consider $$|x(t)| \le \sqrt{t^2+1}$$ for all $$t$$.

If $$|x(t_0)| \le \sqrt{t_0^2+1}$$ at some $$t_0$$, then at that point:
\[
x' = \frac{x^2 - (t^2+1)}{t^2+1} \le 0
\]
So $$|x(t)|$$ cannot increase beyond $$\sqrt{t^2+1}$$.

**Step 3: Initial condition $$|u(0)| < 1$$**
At $$t=0$$, $$\sqrt{0^2+1} = 1$$, so $$|u(0)| < 1$$ means $$|u(0)| < \sqrt{0^2+1}$$.

**Step 4: Proof by contradiction**
Assume solution $$u(t)$$ becomes undefined at finite $$t = T$$. Then it must blow up: $$|u(t)| \to \infty$$ as $$t \to T^-$$.

But for $$|u(t)| > \sqrt{t^2+1}$$, we have $$u' > 0$$ if $$u>0$$, or $$u' < 0$$ if $$u<0$$.

Actually, consider $$V(t) = u(t)^2 - (t^2+1)$$.
\[
V' = 2uu' - 2t = 2u\left(\frac{u^2}{t^2+1} - 1\right) - 2t
\]
This is complicated; better approach:

**Step 5: Simpler approach: Comparison**
Note that $$\dfrac{x^2}{t^2+1} - 1 \le \dfrac{x^2}{1} - 1 = x^2 - 1$$ for all $$t$$.

Compare with $$y' = y^2 - 1$$, which has solution $$y(t) = \tanh(t+C)$$ or $$\coth(t+C)$$.

For $$|y(0)| < 1$$, solution is $$y(t) = \tanh(t+C)$$ which exists for all $$t$$.

Since $$x' \le x^2 - 1$$, by comparison theorem, $$x(t)$$ grows slower than solution to $$y' = y^2 - 1$$.

Thus $$x(t)$$ exists for all $$t$$ if $$|x(0)| < 1$$.

### (c) Isocline for $$m = \frac{1+\sqrt{5}}{2}$$

**Step 1: Find the isocline**
Let $$m = \frac{1+\sqrt{5}}{2} \approx 1.618$$.
\[
x^2 = (m+1)(t^2+1) = \left(\frac{3+\sqrt{5}}{2}\right)(t^2+1)
\]
So $$x = \pm \sqrt{\frac{3+\sqrt{5}}{2}} \sqrt{t^2+1}$$.

**Step 2: Asymptotes for large $$|t|$$**
As $$|t| \to \infty$$, $$\sqrt{t^2+1} \sim |t|$$, so:
\[
x \sim \pm \sqrt{\frac{3+\sqrt{5}}{2}} |t| = \pm k|t| \quad \text{where } k = \sqrt{\frac{3+\sqrt{5}}{2}}
\]
Simplify: $$k^2 = \frac{3+\sqrt{5}}{2}$$.

Note that $$\frac{3+\sqrt{5}}{2} = \left(\frac{1+\sqrt{5}}{2}\right)^2 = \phi^2$$ where $$\phi = \frac{1+\sqrt{5}}{2}$$.

Thus $$k = \phi$$, and asymptotes are $$x = \pm \phi t$$.

**Step 3: Antifunnel between isocline and asymptote**
Consider the region between $$x = \phi t$$ (asymptote) and $$x = \phi\sqrt{t^2+1}$$ (isocline).

For large $$t$$, $$\phi\sqrt{t^2+1} = \phi t\sqrt{1+\frac{1}{t^2}} \approx \phi t\left(1 + \frac{1}{2t^2}\right) = \phi t + \frac{\phi}{2t}$$.

The width of the region is $$\sim \frac{\phi}{2t} \to 0$$ as $$t \to \infty$$.

**Step 4: Verify antifunnel conditions**
An antifunnel requires:

1. Region narrows as $$t \to \infty$$
2. $$f(t,x) = \dfrac{x^2}{t^2+1} - 1$$ is increasing in $$x$$ in the region

Check monotonicity: $$\dfrac{\partial f}{\partial x} = \dfrac{2x}{t^2+1} > 0$$ for $$x > 0$$.

For $$x$$ between $$\phi t$$ and $$\phi\sqrt{t^2+1}$$, certainly $$x > 0$$ for large $$t$$.

Thus we have an antifunnel.

**Step 5: Number of solutions in antifunnel**
By the Antifunnel Theorem, there is at most one solution in an antifunnel. Since the region narrows to zero width, there is exactly one solution in this antifunnel.

### Summary

1. **(a)** Isoclines are hyperbola-like: $$x = \pm\sqrt{m+1}\sqrt{t^2+1}$$; $$m=0$$ gives $$x=\pm\sqrt{t^2+1}$$, $$m=-1$$ gives $$x=0$$.
2. **(b)** Solutions with $$|u(0)|<1$$ exist for all $$t$$ by comparison with $$y'=y^2-1$$.
3. **(c)** Asymptotes: $$x=\pm\phi t$$; there is exactly one solution in the antifunnel.

\[
\boxed{\text{(a) Isoclines: } x=\pm\sqrt{m+1}\sqrt{t^2+1};\quad \text{(b) Solutions exist globally};\quad \text{(c) 1 solution in antifunnel}}
\]

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Problem #14